Probability

arorag · Posted: Sun Jun 29, 2008 5:36 pm Post subject: Probability

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?
ANS 4/9

punit.kaur.mba · Posted: Tue Jul 01, 2008 8:10 am Post subject:

According to me the answer should be 9/15 = 3/5

4/9 looks wrong to me. Reason being, he is simple taking 2/3 * 2/3 = 4/9

Doesnt make sense...

Here is how I get 3/5 -
Denominator - Total no of ways to get 4 children = 6C4
Numerator - Total no of ways to select 2 girls of 3 * Total no of ways to select 2 boys of 3
=(3C2 * 3C2)/6C4
=9/15 = 3/5

arorag, could you please check again. I doubt the OA. Whats the source of the quest..

Anybody has any other explainations??

Ian Stewart · Posted: Tue Jul 01, 2008 10:51 am Post subject:

Yes, the OA (4/9) can't be right here. The answer is 3/5; I used the same approach as in the above post.

beeparoo · Posted: Wed Jul 02, 2008 5:28 pm Post subject:

somail · Posted: Thu Jul 03, 2008 10:57 am Post subject:

Hi,

For some reason I am having a hard time understanding the probablity questions that are usually posted. Can one of you solve this in a little more detail (dumbed down).

Thanks for any help

pranavc · Posted: Thu Jul 03, 2008 6:38 pm Post subject:

I'll give this a shot. You have three guys and three girls to select from. You have 4 spots. The order in these 4 spots is insignificant. So you have to use the formula of combinations to determine the total number of selection possibilities. This boils down to 6C4 (i.e. six people, 4 spots, and order does not matter). This gives you a total of 15 possibilities.

Now, think of the number of ways in which the number of girls will NOT be equal to the number of guys. Two scenarios will results in this.

Scenario 1: 3 girls are selected and one boy is selected

The number of selections that would fit this scenario = 3

because one of three boys can be selected as the "one boy with the three girls".

Scenario 2: 3 boys are selected and one girl is selected

The number of selections that would fit this scenario = 3

because one of three girls can be selected as the "one girl with the three boys".

Hence, the total number of selections in which the number of boys is not equal to the number of girls is 6.

Hence, the number of selections in which the number of boys IS equal to the nunber of girls = 15 (i.e. total number of selection possibilities) - 6

= 9

Hence, the probability of this equal situation = 9/15

I hope this makes sense.

somail · Posted: Mon Jul 07, 2008 12:22 pm Post subject:

Thanks, that made a lot of sense. My biggest problem was that I could not figure out how you guys were getting 15. After you mentioned the "formula of combinations", I googled it and found the equation. The problem was much easier after that.